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An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2R$ from the centre of the earth. What is the gain in its potential energy ?
Solution
Potential energy of the object at the surface of the earth $=-\frac{\mathrm{GM} m}{\mathrm{R}}$
$PE$ of the object at a height equal to the radius of the earth $=-\frac{\mathrm{GM} m}{2 \mathrm{R}}$
$\therefore$ Gain in $PE$ of the object
$=\frac{-\mathrm{GM} m}{2 \mathrm{R}}-\left(-\frac{\mathrm{GM} m}{\mathrm{R}}\right)$
$=\frac{-\mathrm{GM} m+2 \mathrm{GM} m}{2 \mathrm{R}}=+\frac{\mathrm{GM} m}{2 \mathrm{R}}$
$=\frac{g \mathrm{R}^{2} \times m}{2 \mathrm{R}}=\frac{1}{2} m g \mathrm{R} \quad\left(\because g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$cause
$\left(\because g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$
